Wikipedia:Reference desk/Archives/Mathematics/2018 February 6

= February 6 =

Bring radicals
Bring radicals are used to solve quintic equations. BR($$a$$) is defined as the unique real solution of the equation $$x^5+x+a=0.$$ The following suggestions for the article were made on the talk page in October of 2015, but no one has carried them out yet:


 * Show a table and/or graph of the function. (And then of its derivative, integral, etc.)
 * Show expressions for $$\operatorname{BR}(a+b)$$, $$\operatorname{BR}(ab)$$, $$\operatorname{BR}(a^b)$$, etc., or the impossibility thereof.

Does anyone have a reference for either of these? Loraof (talk) 21:21, 6 February 2018 (UTC)

A table
7j2":(,.~0 _1 0 0 0 _1&p.)-0.2*i:10 _34.00  2.00 _20.70   1.80 _12.09   1.60  _6.78   1.40  _3.69   1.20  _2.00   1.00  _1.13   0.80  _0.68   0.60  _0.41   0.40  _0.20   0.20   0.00   0.00   0.20  _0.20   0.41  _0.40   0.68  _0.60   1.13  _0.80   2.00  _1.00   3.69  _1.20   6.78  _1.40  12.09  _1.60  20.70  _1.80  34.00  _2.00 Bo Jacoby (talk) 07:48, 7 February 2018 (UTC)


 * ... and the graph of $$y=\operatorname{BR}(x)$$ is the graph of $$y=x^5+x$$ rotated anticlockwise by 90 degrees. Gandalf61 (talk) 09:34, 7 February 2018 (UTC)

Thanks, both of you! Is there anything about the second bulleted question? I’m guessing that at best it’s like the function $$\sqrt x$$ in the sense that $$\sqrt{ab}$$ can be written in terms of $$\sqrt a$$ and $$\sqrt b, $$ but $$\sqrt{a+b}$$ cannot be. Loraof (talk) 16:13, 7 February 2018 (UTC)
 * Sounds dubious to me: exponentiation (including taking the square root) is distributive over multiplication, but "Bring radicaling" is not (otherwise, you could reconstruct all Bring radicals from BR(2), and I am pretty sure this would be known and mentioned in the article).
 * Probably an expert of the Abel-Ruffini theorem could say more - I have a feeling but no proof that any low-degree polynomial relation between BR(a) and BR(b) (for any a,b) would violate it (general idea: craft a polynomial of degree 5 with a known rational root and which can be reduced to a "Bring quintic", that gives you BR(a) (= the rational root) for some a (which is going to depend on the polynomial reduction process but that involves only an algebraic solution), derive BR(b) for all/all but a countable number of b from the polynomial relation, and you solved the fifth degree by radicals, which should be impossible). Tigraan Click here to contact me 13:15, 9 February 2018 (UTC)